Mathematics Asked by pencil321 on January 3, 2022

Let $f$ be a holomorphic function at the unit circular disk and that $f(0)=0$. Now let $f$ be real valued function on the radii $[0,1)$ and $[0,e^{ifrac{pi}{4}})$ Show that $f$ has a zero of, at least, degree $4$ in $z = 0$

I have no clue where to start!

Thanks in advance!

While @Robert answer suggests how to do the problem using the Taylor series and a local approach, let me show how to do it using the identity principle:

let $f_1(z)=overline {f(bar z)}-f(z)$ which is analytic in the disc;

since $f_1(r)=0, 0 le r <1$ it follows $f_1(z)=0, z in mathbb D$ so $overline {f(bar z)}=f(z)$

and in particular $f(re^{-ipi/4})=overline {f(re^{ipi/4})}=f(re^{ipi/4}), 0 le r <1$

Let $f_2(z)=f(iz)-f(z)$

Since $f_2(re^{-ipi/4})=f(ire^{-ipi/4})-f(re^{-ipi/4})=f(re^{ipi/4})-f(re^{-ipi/4})=0, 0 le r<1$ it follows as before $f_2(z)=0$ in the disc or $f(iz)=f(z)$

Conjugating (or using $f_3(z)=f(-iz)-f(z)$ zero for $z=re^{ipi/4}$) we get $f(-iz)=f(z)$ so in particular $f(iz)=f(-iz)$ hence using $w=iz$ we get $f(w)=f(-w)$ too.

But now the symmetries above mean that $f(z)=frac{f(z)+f(-z)+f(iz)+f(-iz)}{4}=g(z^4)$ for some analytic $g$ in the unit disc

(if $f(z)=sum a_kz^k$ then $g(z)=sum a_{4k}z^k$)

This immediately implies that if $f$ has a zero at $0$ it has a zero of order $4k, k ge 1$ so the order is at least $4$ and more generally the next value possible is $8$ etc

Answered by Conrad on January 3, 2022

Hint: suppose the zero at $z=0$ has degree $k$. What does the Maclaurin series of $f(z)$ look like? What does that say about $f(z)$ when $z$ is close to $0$ and is on one of those two radii?

Answered by Robert Israel on January 3, 2022

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